${\sqrt[3]{7290} = \text{?}}$
Solution: $\sqrt[3]{7290}$ is the number that, when multiplied by itself three times, equals $7290$ First break down $7290$ into its prime factorization and look for factors that appear three times. So the prime factorization of $7290$ is $2\times 3\times 3\times 3\times 3\times 3\times 3\times 5$ Notice that we can rearrange the factors like so: $7290 = 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 = (3\times 3\times 3) \times (3\times 3\times 3) \times 2\times 5$ So $\sqrt[3]{7290} = \sqrt[3]{3\times 3\times 3} \times \sqrt[3]{3\times 3\times 3} \times \sqrt[3]{2\times 5}$ $\sqrt[3]{7290} = 3\times 3 \times \sqrt[3]{2\times 5}$ $\sqrt[3]{7290} = 9 \sqrt[3]{10}$